Problem: Simplify and expand the following expression: $ \dfrac{3}{2n - 6}- \dfrac{3}{n + 9}- \dfrac{4}{n^2 + 6n - 27} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $2$ out of denominator in the first term: $ \dfrac{3}{2n - 6} = \dfrac{3}{2(n - 3)}$ We can factor the quadratic in the third term: $ \dfrac{4}{n^2 + 6n - 27} = \dfrac{4}{(n - 3)(n + 9)}$ Now we have: $ \dfrac{3}{2(n - 3)}- \dfrac{3}{n + 9}- \dfrac{4}{(n - 3)(n + 9)} $ The least common multiple of the denominators is: $ 2(n - 3)(n + 9)$ In order to get the first term over $2(n - 3)(n + 9)$ , multiply by $\dfrac{n + 9}{n + 9}$ $ \dfrac{3}{2(n - 3)} \times \dfrac{n + 9}{n + 9} = \dfrac{3(n + 9)}{2(n - 3)(n + 9)} $ In order to get the second term over $2(n - 3)(n + 9)$ , multiply by $\dfrac{2(n - 3)}{2(n - 3)}$ $ \dfrac{3}{n + 9} \times \dfrac{2(n - 3)}{2(n - 3)} = \dfrac{6(n - 3)}{2(n - 3)(n + 9)} $ In order to get the third term over $2(n - 3)(n + 9)$ , multiply by $\dfrac{2}{2}$ $ \dfrac{4}{(n - 3)(n + 9)} \times \dfrac{2}{2} = \dfrac{8}{2(n - 3)(n + 9)} $ Now we have: $ \dfrac{3(n + 9)}{2(n - 3)(n + 9)} - \dfrac{6(n - 3)}{2(n - 3)(n + 9)} - \dfrac{8}{2(n - 3)(n + 9)} $ $ = \dfrac{ 3(n + 9) - 6(n - 3) - 8} {2(n - 3)(n + 9)} $ Expand: $ = \dfrac{3n + 27 - 6n + 18 - 8}{2n^2 + 12n - 54} $ $ = \dfrac{-3n + 37}{2n^2 + 12n - 54}$